SelectMany for flattening in C Sharp (C#) - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to flatten the list of lists using LINQ.

C Sharp (C#)

var numbers = new List<List<int>> { new List<int> {1, 2}, new List<int> {3, 4} };
var flat = numbers.[1](x => x).ToList();

Drag options to blanks, or click blank then click option'

AOrderBy

BSelect

CSelectMany

DWhere

Attempts:

3 left

💡 Hint

Common Mistakes

Using Select instead of SelectMany returns a list of lists, not a flat list.

The SelectMany method flattens a collection of collections into a single collection.

2fill in blank

medium

Complete the code to flatten and filter numbers greater than 2.

C Sharp (C#)

var numbers = new List<List<int>> { new List<int> {1, 2}, new List<int> {3, 4} };
var result = numbers.[1](x => x).Where(n => n > 2).ToList();

Drag options to blanks, or click blank then click option'

AOrderBy

BSelect

CWhere

DSelectMany

Attempts:

3 left

💡 Hint

Common Mistakes

Using Select only returns nested lists, so Where filters lists, not numbers.

SelectMany flattens the list, then Where filters numbers greater than 2.

3fill in blank

hard

Fix the error in the code to flatten the list of arrays.

C Sharp (C#)

string[][] words = { new string[] {"apple", "banana"}, new string[] {"cherry"} };
var flatWords = words.[1](w => w).ToList();

Drag options to blanks, or click blank then click option'

ASelectMany

BSelect

CWhere

DOrderBy

Attempts:

3 left

💡 Hint

Common Mistakes

Using Select causes flatWords to be a list of arrays, not strings.

SelectMany flattens the array of arrays into a single list of strings.

4fill in blank

hard

Fill both blanks to flatten and select the length of each word.

C Sharp (C#)

var words = new List<List<string>> { new List<string> {"cat", "dog"}, new List<string> {"bird"} };
var lengths = words.[1](w => w).Select(word => word.[2]).ToList();

Drag options to blanks, or click blank then click option'

ASelectMany

BSelect

CLength

DCount

Attempts:

3 left

💡 Hint

Common Mistakes

Using Select instead of SelectMany keeps nested lists.

Using Count instead of Length on strings causes errors.

SelectMany flattens the list, then Length gets each word's length.

5fill in blank

hard

Fill all three blanks to flatten, filter words longer than 3, and convert to uppercase.

C Sharp (C#)

var words = new List<List<string>> { new List<string> {"cat", "dog"}, new List<string> {"bird", "fish"} };
var result = words.[1](w => w).Where(word => word.[2] > 3).Select(word => word.[3]()).ToList();

Drag options to blanks, or click blank then click option'

ASelect

BLength

CToUpper

DSelectMany

Attempts:

3 left

💡 Hint

Common Mistakes

Using Select instead of SelectMany keeps nested lists.

Using Count instead of Length on strings causes errors.

Forgetting parentheses on ToUpper causes errors.

SelectMany flattens the list, Length filters words longer than 3, and ToUpper() converts words to uppercase.

Practice

1. What does the SelectMany method do in C# when used on a list of lists?

easy

A. It sorts each inner list individually.

B. It filters the outer list based on a condition.

C. It flattens the nested lists into a single list containing all elements.

D. It creates a dictionary from the nested lists.

Solution

1. Step 1: Understand the structure of nested lists

   A list of lists means you have many small lists inside one big list.

2. Step 2: Role of SelectMany

   SelectMany takes each small list and combines all their elements into one big flat list.

3. Final Answer:

   It flattens the nested lists into a single list containing all elements. -> Option C

4. Quick Check:

   SelectMany flattens nested lists = B [OK]

Hint: SelectMany flattens nested lists into one list fast [OK]

Common Mistakes:

• Thinking SelectMany sorts or filters instead of flattening
• Confusing SelectMany with Select which keeps nested structure
• Assuming SelectMany creates dictionaries

2. Which of the following is the correct syntax to flatten a list of lists List numbers using SelectMany?

easy

A. var flat = numbers.SelectMany(x => x);

B. var flat = numbers.Select(x => x);

C. var flat = numbers.SelectMany(x => x.ToString());

D. var flat = numbers.SelectMany(x => x.Count());

Solution

1. Step 1: Understand SelectMany syntax

   SelectMany expects a function that returns an inner collection to flatten.

2. Step 2: Check each option

   var flat = numbers.SelectMany(x => x); returns each inner list as is, which is correct. var flat = numbers.Select(x => x); uses Select, not flattening. Options C and D return wrong types (string and int), not collections.

3. Final Answer:

   var flat = numbers.SelectMany(x => x); -> Option A

4. Quick Check:

   SelectMany with identity function flattens = A [OK]

Hint: Use SelectMany with identity function to flatten lists [OK]

Common Mistakes:

• Using Select instead of SelectMany
• Returning non-collections inside SelectMany
• Trying to flatten by counting elements

3. What is the output of the following code?

var data = new List {
  new List<int> {1, 2},
  new List<int> {3, 4},
  new List<int> {5}
};
var result = data.SelectMany(x => x).ToList();
Console.WriteLine(string.Join(",", result));

medium

A. 12345

B. 1,2;3,4;5

C. System.Collections.Generic.List`1[System.Int32]

D. 1,2,3,4,5

Solution

1. Step 1: Understand SelectMany flattening

   SelectMany flattens all inner lists into one list: [1,2,3,4,5].

2. Step 2: Output formatting

   string.Join with "," joins elements with commas, so output is "1,2,3,4,5".

3. Final Answer:

   1,2,3,4,5 -> Option D

4. Quick Check:

   Flattened list joined by commas = D [OK]

Hint: SelectMany flattens, string.Join prints with commas [OK]

Common Mistakes:

• Expecting nested list output instead of flat
• Confusing ToString output of list object
• Missing commas in output

4. Identify the error in this code snippet that tries to flatten a list of lists:

List words = new() {
  new List<string> {"apple", "banana"},
  new List<string> {"cat", "dog"}
};
var flatWords = words.SelectMany(w => w.Length).ToList();

medium

A. The list initialization syntax is incorrect.

B. Using w.Length returns int, not a collection, causing a compile error.

C. Missing parentheses after SelectMany call.

D. ToList() cannot be called after SelectMany.

Solution

1. Step 1: Analyze the lambda inside SelectMany

   The lambda uses w => w.Length, which returns an int, not a collection.

2. Step 2: Understand SelectMany requirements

   SelectMany expects a function returning a collection to flatten, not a single int value.

3. Final Answer:

   Using w.Length returns int, not a collection, causing a compile error. -> Option B

4. Quick Check:

   SelectMany needs collections, not int = C [OK]

Hint: SelectMany lambda must return collections, not single values [OK]

Common Mistakes:

• Returning single values instead of collections in SelectMany
• Confusing Length property with collection
• Assuming ToList() causes error

5. Given a list of students where each student has a list of grades, how would you use SelectMany to get a list of all grades above 80 from all students?

class Student {
  public string Name { get; set; }
  public List<int> Grades { get; set; }
}

List<Student> students = new() {
  new Student { Name = "Anna", Grades = new List<int> { 90, 70 } },
  new Student { Name = "Ben", Grades = new List<int> { 85, 60 } }
};

// Fill in the blank to get all grades > 80
var highGrades = students.SelectMany(__________).ToList();

hard

A. s => s.Grades.Where(g => g > 80)

B. s => s.Grades

C. s => s.Name

D. s => s.Grades.Count

Solution

1. Step 1: Understand the goal

   We want all grades above 80 from all students combined.

2. Step 2: Use SelectMany with filtering inside

   s => s.Grades.Where(g => g > 80) filters grades > 80 inside SelectMany, flattening only those grades.

3. Step 3: Why not others?

   s => s.Grades flattens all grades, including low ones like 70 and 60. Options C and D return wrong types (string and int).

4. Final Answer:

   s => s.Grades.Where(g => g > 80) -> Option A

5. Quick Check:

   SelectMany with filtered inner lists = A [OK]

Hint: Filter inside SelectMany to flatten only needed items [OK]

Common Mistakes:

• Flattening all grades without filtering
• Returning non-collections in SelectMany
• Using wrong property like Name or Count